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Problems
Contests
National and Regional Contests
Canada Contests
Canada National Olympiad
2022 Canada National Olympiad
2022 Canada National Olympiad
Part of
Canada National Olympiad
Subcontests
(5)
5
1
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Inscribed/circumscribed pentagon
A pentagon is inscribed in a circle, such that the pentagon has an incircle. All
10
10
10
sets of
3
3
3
vertices from the pentagon are chosen, and the incenters of each of the
10
10
10
resulting triangles are drawn in. Prove these
10
10
10
incenters lie on
2
2
2
concentric circles.Note: I spent nearly no time on this, so if anyone took CMO and I misremembered just let me know.
4
1
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Colourable combo geo
Call a set of
n
n
n
lines good if no
3
3
3
lines are concurrent. These
n
n
n
lines divide the Euclidean plane into regions (possible unbounded). A coloring is an assignment of two colors to each region, one from the set
{
A
1
,
A
2
}
\{A_1, A_2\}
{
A
1
,
A
2
}
and the other from
{
B
1
,
B
2
,
B
3
}
\{B_1, B_2, B_3\}
{
B
1
,
B
2
,
B
3
}
, such that no two adjacent regions (adjacent meaning sharing an edge) have the same
A
i
A_i
A
i
color or the same
B
i
B_i
B
i
color, and there is a region colored
A
i
,
B
j
A_i, B_j
A
i
,
B
j
for any combination of
A
i
,
B
j
A_i, B_j
A
i
,
B
j
.A number
n
n
n
is colourable if there is a coloring for any set of
n
n
n
good lines. Find all colourable
n
n
n
.
3
1
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Process combo
Vishal starts with
n
n
n
copies of the number
1
1
1
written on the board. Every minute, he takes two numbers
a
,
b
a, b
a
,
b
and replaces them with either
a
+
b
a+b
a
+
b
or
min
(
a
2
,
b
2
)
\min(a^2, b^2)
min
(
a
2
,
b
2
)
. After
n
−
1
n-1
n
−
1
there is
1
1
1
number on the board. Let the maximal possible value of this number be
f
(
n
)
f(n)
f
(
n
)
. Prove
2
n
/
3
<
f
(
n
)
≤
3
n
/
3
2^{n/3}<f(n)\leq 3^{n/3}
2
n
/3
<
f
(
n
)
≤
3
n
/3
.
1
1
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symmetric algebra
If
a
b
+
a
b
+
1
+
a
2
+
b
a
+
b
2
=
0
ab+\sqrt{ab+1}+\sqrt{a^2+b}\sqrt{a+b^2}=0
ab
+
ab
+
1
+
a
2
+
b
a
+
b
2
=
0
, find the value of
b
a
2
+
b
+
a
b
2
+
a
b\sqrt{a^2+b}+a\sqrt{b^2+a}
b
a
2
+
b
+
a
b
2
+
a
2
1
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number theory
I think we are allowed to discuss since its after 24 hours How do you do this Prove that
d
(
1
)
+
d
(
3
)
+
.
.
+
d
(
2
n
−
1
)
≤
d
(
2
)
+
d
(
4
)
+
.
.
.
d
(
2
n
)
d(1)+d(3)+..+d(2n-1)\leq d(2)+d(4)+...d(2n)
d
(
1
)
+
d
(
3
)
+
..
+
d
(
2
n
−
1
)
≤
d
(
2
)
+
d
(
4
)
+
...
d
(
2
n
)
which
d
(
x
)
d(x)
d
(
x
)
is the divisor function